Prove $\int_0^1|f(t)-g(t)|dt \le (\int_0^1|f(t)-g(t)|^2dt)^{1/2} \le \sup_{t\in[0,1]}|f(t)-g(t)|$ Let $C[0,1]$ be the set of all continuous real-valued functions on $[0,1]$.
Let these be 3 metrics on $C$.
$p(f,g)=\sup_{t\in[0,1]}|f(t)-g(t)|$
$d(f,g)=(\int_0^1|f(t)-g(t)|^2dt)^{1/2}$
$t(f,g)=\int_0^1|f(t)-g(t)|dt$
Prove that for every $f,g\in C$, the following holds $t(f,g)\le d(f,g)\le p(f,g)$
I understand that $t(f,g)\le p(f,g)$ since $t(f,g)=\int_0^1|f(t)-g(t)|dt \le \int_0^1\sup_{t\in[0,1]}|f(t)-g(t)|dt =\sup_{t\in[0,1]}|f(t)-g(t)|=p(f,g)$.
But I can't get the others.
I think using Schwarz's inequality might be useful $(\int_0^1 w(t)v(t)dt)^2 \le (\int_0^1w^2(t)dt)(\int_0^1v^2(t)dt)$
 A: You can adapt your proof that $t(f,g) \leq p(f,g)$ to show $d(f,g) \leq p(f,g)$.
For $t(f,g) \leq d(f,g)$, use Cauchy-Schwarz on
$$
\int_0^1 \left|h\right| \,dx = \int_0^1 \left|h\right| \cdot 1 \,dx
$$
where $h = f - g$.

 In case you don't see it, the integral above is bounded by $||h||_{2} ||1||_{2}$

A: The same reasoning you make you can apply to show $d(f,g)\leq p(f,g)$. So that leaves you with $t(f,g)\leq d(f,g)$. 
This last one is a particular case of the Cauchy-Schwarz Inequality. You start with the number inequality
$$
|ab|\leq\frac{|a|^2}2+\frac{|b|^2}2.
$$
Applied to an integral you get
$$
\int_0^1|h(t)k(t)|\,dt\leq\frac12\,\int_0^1|h(t)|^2\,dt+\frac12\,\int_0^1|k(t)|^2\,dt.
$$
In particular, you get
$$
\int_0^1\left|\frac{h(t)}{\left(\int_0^1|h(t)^2\,dt\right)^{1/2}}\,\frac{k(t)}{\left(\int_0^1|k(t)^2\,dt\right)^{1/2}}\right|\leq1,
$$
or
$$
\int_0^1|h(t)k(t)|\,dt\leq\left(\int_0^1|h(t)|^2\,dt\right)^{1/2}\,\left(\int_0^1|k(t)|^2\,dt\right)^{1/2}.
$$
Now you take $h(t)=f(t)-g(t)$, $k(t)=1$, to get $t(f,g)\leq d(f,g)$.
